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Complex roots: the roots of unity FP2 question watch

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    Hello. I wonder if anyone can explain why the answer to the question is so:

    Q) A regular hexagon is inscribed in the unit circle. One vertex is α. Give the other vertices in terms of α and ω , where ω is a complex cube root of unity.

    A) -α; ±αω, ±αω^2
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    Zacken B_9710



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    I know it should be fairly obvious, but really struggling to get roots of unity - can anyone give me a hand?
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    (Original post by MadeOnTheTrack)
    I know it should be fairly obvious, but really struggling to get roots of unity - can anyone give me a hand?
    Do you know what the sixth roots of unity are in terms of the cube roots of unity?

    Also do you know what happens in the Argand diagram if you multiply a complex number by another complex number?
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    (Original post by 16Characters....)
    Do you know what the sixth roots of unity are in terms of the cube roots of unity?

    Also do you know what happens in the Argand diagram if you multiply a complex number by another complex number?
    Right, I know the second point (it's spiral dilatation) but not the first.
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    (Original post by MadeOnTheTrack)
    Right, I know the second point (it's spiral dilatation) but not the first.
    This is FP2? Edexcel?
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    (Original post by MadeOnTheTrack)
    Right, I know the second point (it's spiral dilatation) but not the first.
    The set of sixth roots of unity is the set of the cube roots of unity + the set of their negatives.

    If you are with OCR MEI (which I think you are, I remember this question) have a look at the first question of this exercise, and replace 10 with 6 :-)
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    (Original post by MadeOnTheTrack)
    Right, I know the second point (it's spiral dilatation) but not the first.
    Hold on, it's just clicked in my head!
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    Yep it's OCR MEI. Thanks for your help at this time of night!
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    Hi guys, it's me again! My answer to this question is wrong but not sure where the fault is in my working:

    Q) Solve the equation (z + j)^n + (z - j)^n = 0.

    A) cot ((2k+1)pi)/2n) , k=0,1,2,…,n-1

    My working and answer is shown in the photo (please ignore first one rotated sideways)
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